RE: [PHP-DB] Re: PHP advice

You should enclose any string that you searching on within a ' and ' Your SQL query should look something like; $query = "SELECT * FROM auth WHERE password = '". $password ."'"; Cheers "Greg Lobring" <globring@hotmail.com> wrote in message news:F116N28f45wSz6btzsY000059bb@hotmail.com... > I dont understand why this is causing me so much trouble, hopefully someone > can help. I have a simple PHP script that takes the user id and password > from a form on a page...here is the php used after submitting the form: > > <? > @ $db = mysql_pconnect("localhost", "blah", "blah"); > if (!$db) > { > echo "Error: Could not connect to database."; > exit; > } > mysql_select_db("blahblah"); > $query = "select * from auth where password = ".$password.""; > $result = mysql_query($query); > $num_results = mysql_num_rows($result); > echo "Number of Rows: ".$num_results.""; > mysql_close($db); > ?> > > It connects fine, and when the query returns NO (0) rows, it appears > correct. However, when the query returns a record, I get the following > error: > > > Warning: Supplied argument is not a valid MySQL result resource in picks.php > on line 16 > > Where line 16 is the the line that says $num_results = > mysql_num_rows($result); > > > Does anyone have any advice? > > Appreciated in advance.. > > Greg > > > > _________________________________________________________________ > Chat with friends online, try MSN Messenger: > http://messenger.msn.com > -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php